How to show the smooth map $f : T^2 \to S^3$ is an orientation-preserving diffeomorphism?

Question

Let $f : T^2 \to S^3$ be the smooth map of a 2-torus into $S^3$. Let $\omega$ be a closed 2-form on $S^3$. Show that $$\int_{T^2}f^*\omega = 0.$$

So apparently, if I can use the theorem on Guillemin and Pollack's Differential Topology, so I change $\int_{T^2}f^*\omega$ to $\int_{S^3}\omega$ then I can prove it by Stokes Theorem.

Theorem [Page 168] If $f: Y \to X$ is an orientation-preserving diffeomorphism, then $$\int_X \omega = \int_Y f^* \omega$$ for every compactly supported, smooth $k$-form on $X$ ($k = \dim X = \dim Y$).

However, the problem is, how can I justify $f$ is an orientation-preserving diffeomorphism?

Thank you. :-)

$T^2$ is $2$-dimensional and $S^3$ is $3$-dimensional, so how can $T^2$ and $S^3$ possibly be homeomorphic, let alone diffeomorphic via an orientation-preserving diffeomorphism? — Branimir Ćaćić, Jul 30 '13 at 18:30
Oops, seems I am trying to use the wrong theorem..@BranimirĆaćić — WishingFish, Jul 30 '13 at 18:31

Branimir Ćaćić · Accepted Answer · 2013-08-06T20:43:42.600

The first fact you need to know is that every closed $2$-form on $S^3$ is exact. If you're happy with algebraic topology, then this is precisely the fact that $H^2_{\text{dR}}(S^3) = 0$. Hence, $\omega = d\eta$ for some $1$-form $\eta$ on $S^3$.

The second fact you need to know is that taking pullbacks commutes with taking exterior derivatives. Hence, $$ f^\ast \omega = f^\ast d \eta = d f^\ast \eta. $$

Now, applying this last observation, observe that $$ \int_{T^2} f^\ast \omega = \int_{T^2} d f^\ast \eta. $$ So, what is $\partial T^2$, and what does this imply about $\int_{T^2} d f^\ast \eta$?

How to show the smooth map $f : T^2 \to S^3$ is an orientation-preserving diffeomorphism?

1 Answers1